1. **State the problem:** Show that the equation $$5x^2 + 5y^2 + 24x + 36y + 10 = 0$$ represents a circle and find its centre and radius. 2. **Rewrite the equation in standard form:** Divide the entire equation by 5 to simplify the coefficients of $x^2$ and $y^2$ to 1. $$\cancel{5}x^2 + \cancel{5}y^2 + \frac{24}{\cancel{5}}x + \frac{36}{\cancel{5}}y + \frac{10}{\cancel{5}} = 0 \Rightarrow x^2 + y^2 + \frac{24}{5}x + \frac{36}{5}y + 2 = 0$$ 3. **Compare with the general form of a circle:** The general form is $$x^2 + y^2 + 2gx + 2fy + c = 0$$ From the equation, we identify: $$2g = \frac{24}{5} \Rightarrow g = \frac{12}{5}$$ $$2f = \frac{36}{5} \Rightarrow f = \frac{18}{5}$$ $$c = 2$$ 4. **Find the centre:** The centre of the circle is $$(-g, -f) = \left(-\frac{12}{5}, -\frac{18}{5}\right)$$ 5. **Find the radius:** The radius is given by $$r = \sqrt{g^2 + f^2 - c}$$ Calculate inside the square root: $$g^2 = \left(\frac{12}{5}\right)^2 = \frac{144}{25}$$ $$f^2 = \left(\frac{18}{5}\right)^2 = \frac{324}{25}$$ $$g^2 + f^2 - c = \frac{144}{25} + \frac{324}{25} - 2 = \frac{144 + 324 - 50}{25} = \frac{418}{25}$$ Therefore, $$r = \sqrt{\frac{418}{25}} = \frac{\sqrt{418}}{5}$$ **Final answer:** The equation represents a circle with centre $$\left(-\frac{12}{5}, -\frac{18}{5}\right)$$ and radius $$\frac{\sqrt{418}}{5}$$.